[Leetcode]-Pascal's Triangle
阿新 • • 發佈:2017-08-06
reg data- fadein asc hid 生成 num %d 帕斯卡三角形
Given numRows, generate the first numRows of Pascal’s triangle.
For example, given numRows = 5,
Return
[
——-[1],
——[1,1],
—-[1,2,1],
—[1,3,3,1],
–[1,4,6,4,1]
]
題目:依據numRows生成帕斯卡三角形。帕斯卡三角形原理點擊這裏
思路:帕斯卡三角形的每行首、位元素都為1。其它的元素為上一行兩個元素之和:
if(0 == j || i == j) r[i][j] = 1;
else r[i ][j] = r[i-1][j-1] + r[i-1][j];
註意事項:
1、numRows為0的時候返回0
2、一定要自己分配內存空間
#include <stdlib.h>
#include <stdio.h>
/**
* Return an array of arrays of size *returnSize.
* The sizes of the arrays are returned as *columnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
int** generate(int numRows, int** columnSizes, int* returnSize) {
if(0 == numRows) return 0;
*returnSize = numRows ;
int i = 0, j = 0;
int** r = (int**)malloc(sizeof(int*) * numRows);
//int*cn = (int*)malloc(sizeof(int) * numRows);
*columnSizes = (int*)malloc(sizeof(int ) * numRows);
//*columnSizes = cn;
for(i=0;i<numRows;i++)
{
//cn[i] = i + 1;
columnSizes[0][i] = i + 1 ;
r[i] = (int*)malloc(sizeof(int)*(i+1));
}
for(i=0;i<numRows;i++)
{
for(j=0;j<i+1;j++)
{
if(0 == j || i == j) r[i][j] = 1;
else r[i][j] = r[i-1][j-1] + r[i-1][j];
}
}
return r;
}
int main()
{
int num = 3 ;
int *columnSizes ;//每一行有幾個元素
int returnSize = 0;
int **r = generate(num,&columnSizes,&returnSize);
int i = 0;
int j = 0;
printf("toal element size is : %d\n",returnSize);
for(i=0;i<num;i++)
{
printf("%d row have %d element \n",i,columnSizes[i]);
}
for(i=0;i<num;i++)
{
for(j=0;j<i+1;j++)
printf("r[%d][%d] = %d\n",i,j,r[i][j]);
printf("\n");
}
}
[Leetcode]-Pascal's Triangle